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5t^2-39t+22=0
a = 5; b = -39; c = +22;
Δ = b2-4ac
Δ = -392-4·5·22
Δ = 1081
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-\sqrt{1081}}{2*5}=\frac{39-\sqrt{1081}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1081}}{2*5}=\frac{39+\sqrt{1081}}{10} $
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